Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3 ), t = time (min), k = the evaporation rate (mm/min), and A = surface area (mm2 ). Use Euler’s method to compute the volume of the droplet from t = 0 to 10 min using a step size of 0.25 min. Assume that k = 0.08 mm/min and that the droplet initially has a radius of 2.5 mm. Assess the validity of your results by determining the radius of your final computed volume and verifying that it is consistent with the evaporation rate.

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shahnoorazhar3 shahnoorazhar3 · Answer 1 · 2020-01-23T19:26:05+01:00

Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

[tex]\frac{dV}{dt} = -kA[/tex]

Using Euler's method

[tex]\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i} \\[/tex]

Where initial droplet volume is:

[tex]V(0) = \frac{4pi}{3} * r(0)^3 = \frac{4pi}{3} * 2.5^3 = 65.45 mm^3[/tex]

Hence, the iterative solution will be as next:

[tex]V'_{i} = -k *4pi*(\frac{3*65.45}{4pi})^(2/3) = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88[/tex]

[tex]V'_{i} = -k *4pi*(\frac{3*63.88}{4pi})^(2/3) = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33[/tex]

[tex]V'_{i} = -k *4pi*(\frac{3*62.33}{4pi})^(2/3) = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813[/tex]

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

[tex]r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\[/tex]

The average change of droplet radius with time is:

Δr/Δt = [tex]\frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min[/tex]

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!